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Unformatted text preview: mata (jpm2873) – HW #7 – Antoniewicz – (57420) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 1930 kg car starts from rest and accelerates uniformly to 18 . 9 m / s in 19 s . Find the average power developed by the engine. Assume that air resistance remains constant at 397 N during this time. Correct answer: 29 . 3487 hp. Explanation: Let : m = 1930 kg , v i = 0 m / s , v f = 18 . 9 m / s , and Δ t = 19 s . The acceleration of the car is a = v f v i Δ t = v f Δ t = 18 . 9 m / s 19 s = 0 . 994737 m / s 2 and the constant forward force due to the engine is found from summationdisplay F = F engine F air = ma F engine = F air + ma = 397 N + (1930 kg) ( . 994737 m / s 2 ) = 2316 . 84 N . The average velocity of the car during this interval is v av = v f + v i 2 , so the average power output is P = F engine v av = F engine parenleftBig v f 2 parenrightBig = (2316 . 84 N) parenleftbigg 18 . 9 m / s 2 parenrightbiggparenleftbigg 1 hp 746 W parenrightbigg = 29 . 3487 hp . 002 (part 2 of 2) 10.0 points Find the instantaneous power output of the engine at t = 19 s just before the car stops accelerating. Correct answer: 58 . 6975 hp. Explanation: The instantaneous velocity is 18 . 9 m / s and the instantaneous power output of the engine is P = F engine v f = (2316 . 84 N)(18 . 9 m / s) parenleftbigg 1 hp 746 W parenrightbigg = 58 . 6975 hp . 003 (part 1 of 2) 10.0 points On the way to the moon, the Apollo astro nauts reach a point where the Moon’s gravi tational pull is stronger than that of Earth’s. Find the distance of this point from the center of the Earth. The masses of the Earth and the Moon are 5 . 98 × 10 24 kg and 7 . 36 × 10 22 kg, respectively, and the distance from the Earth to the Moon is 3 . 84 × 10 8 m. Correct answer: 3 . 45653 × 10 8 m. Explanation: Let : M e = 5 . 98 × 10 24 kg , m m = 7 . 36 × 10 22 kg , and R e = 3 . 84 × 10 8 m . Consider the point where the two forces are equal. If r e is the distance from this point to the center of the Earth and r m is the distance from this point to the center of the Moon, then GmM e r 2 e = GmM m r 2 m r 2 m r 2 e = M m M e r m r e = radicalbigg M m M e . mata (jpm2873) – HW #7 – Antoniewicz – (57420) 2 It is also true that R = r e + r m = r e + radicalbigg M m M e r e r e = R 1 + radicalbigg M m M e = 3 . 84 × 10 8 m 1 + radicalBigg 7 . 36 × 10 22 kg 5 . 98 × 10 24 kg = 3 . 45653 × 10 8 m . 004 (part 2 of 2) 10.0 points What is the acceleration due to the Earth’s gravity at this point? The value of the universal gravitational constant is 6 . 672 × 10 − 11 N · m 2 / kg 2 ....
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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